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Question
a. prove that a quadrilateral whose diagonals are congruent and bisect each other is a rectangle.
b. explain how to use part (a) and only a compass and straightedge to construct any rectangle.
c. construct another rectangle not congruent to the rectangle in part (b) but whose diagonals are congruent to the diagonals of the rectangle in part (b). why are the rectangles not congruent?
a. let ac and bd be two line - segments that bisect each other at e, with ac≅bd. prove that abcd is a rectangle.
because ac and bd bisect each other, abcd is a parallelogram and its opposite sides are congruent and parallel.
due to sss congruence, △abc≅△dcb.
because corresponding parts of congruent triangles are congruent, it follows that ∠abc≅∠dcb.
because these angles are also
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Step1: Recall parallelogram property
Since diagonals $\overline{AC}$ and $\overline{BD}$ bisect each other, quadrilateral $ABCD$ is a parallelogram. In a parallelogram, opposite sides are congruent and parallel.
Step2: Prove triangle - congruence
In $\triangle ABC$ and $\triangle DCB$, we have $AB = DC$ (opposite sides of parallelogram), $BC=CB$ (common side) and $AC = BD$ (given). By SSS (Side - Side - Side) congruence criterion, $\triangle ABC\cong\triangle DCB$.
Step3: Use corresponding - parts of congruent triangles
Since $\triangle ABC\cong\triangle DCB$, corresponding parts of congruent triangles are congruent, so $\angle ABC\cong\angle DCB$.
Step4: Identify angle - relationship
$\angle ABC$ and $\angle DCB$ are interior angles on the same side of a transversal (in this case, $BC$ is part of a transversal for parallel lines $AB$ and $DC$ in parallelogram $ABCD$). In a parallelogram, adjacent angles are supplementary, i.e., $\angle ABC+\angle DCB = 180^{\circ}$. Since $\angle ABC=\angle DCB$, we have $2\angle ABC=180^{\circ}$, so $\angle ABC = 90^{\circ}$. A parallelogram with one right - angle is a rectangle.
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Interior angles on the same side of a transversal